Question Number 92137 by hore last updated on 05/May/20
$$\int\frac{\mathrm{2}{x}}{\mathrm{1}+{x}} \\ $$
Commented by john santu last updated on 05/May/20
$$\int\:\frac{\mathrm{2}{x}\:{dx}}{{x}+\mathrm{1}}\:=\:\int\:\frac{\mathrm{2}\left({x}+\mathrm{1}\right)−\mathrm{2}}{{x}+\mathrm{1}}\:{dx} \\ $$$$=\:\int\:\mathrm{2}\:−\:\frac{\mathrm{2}}{{x}+\mathrm{1}}\:{dx}\: \\ $$$$=\:\mathrm{2}{x}−\mathrm{2}\:\mathrm{ln}\mid{x}−\mathrm{1}\mid\:+\:\mathrm{c}\: \\ $$
Commented by Rio Michael last updated on 07/May/20
![((2x)/(1 +x)) = 2 −(2/(x+1)) [can use long division] ∫ ((2x)/(1+x)) dx = ∫(2 −(2/(x +1))) dx = 2x− 2 ln(x + 1) + C](Q92473.png)
$$\:\frac{\mathrm{2}{x}}{\mathrm{1}\:+{x}}\:=\:\mathrm{2}\:−\frac{\mathrm{2}}{{x}+\mathrm{1}}\:\left[\mathrm{can}\:\mathrm{use}\:\mathrm{long}\:\mathrm{division}\right] \\ $$$$\int\:\frac{\mathrm{2}{x}}{\mathrm{1}+{x}}\:{dx}\:=\:\int\left(\mathrm{2}\:−\frac{\mathrm{2}}{{x}\:+\mathrm{1}}\right)\:{dx}\:=\:\mathrm{2}{x}−\:\mathrm{2}\:\mathrm{ln}\left({x}\:+\:\mathrm{1}\right)\:+\:{C} \\ $$$$ \\ $$